# Thread: Hi! I am a Computer Science Engineer studying math for an exam

1. ## Hi! I am a Computer Science Engineer studying math for an exam

Hi, i am a Computer Science Engineer.. I'm studying math for an exam and never been good at it.. now trying my best bottom up!! need help in a lot of problems.. the first one is here , thanx a lot!

This is a question from the ISC book M. L. Agarwal. Please tell this approach to this question.
How many words can be made with letters of the word INTERMEDIATE if
(i) the words neither begin with I nor end with E
(ii) the relative order of vowels and consonants does not change.

Ans. (i) 83X((10!)/24) (ii) 21600

2. ## Re: Hi! I am a Computer Science Engineer studying math for an exam

Originally Posted by rachitmagon
Hi, i am a Computer Science Engineer.. I'm studying math for an exam and never been good at it.. now trying my best bottom up!! need help in a lot of problems.. the first one is here , thanx a lot!
This is a question from the ISC book M. L. Agarwal. Please tell this approach to this question.
How many words can be made with letters of the word INTERMEDIATE if
(i) the words neither begin with I nor end with E
(ii) the relative order of vowels and consonants does not change.
Ans. (i) 83X((10!)/24) (ii) 21600

3. ## Re: Hi! I am a Computer Science Engineer studying math for an exam

Originally Posted by Plato
The approach to deal with this question, i know the numeric answer from a source but i don't know how to go about it.

4. ## Re: Hi! I am a Computer Science Engineer studying math for an exam

Originally Posted by rachitmagon
The approach to deal with this question, i know the numeric answer from a source but i don't know how to go about it.
Originally Posted by rachitmagon
How many words can be made with letters of the word INTERMEDIATE if
(i) the words neither begin with I nor end with E
Ans. (i) 83X((10!)/24)
Because of repeated letters the total number of arrangements is $\frac{12!}{(2!)^2(3!)}$.
Now the answer the question you need to find the number of those that begin with I plus the number that end with E minus the number beginning with I and end with E.

So $\frac{11!}{(2!)(3!)}+\frac{11!}{(2!)^3}-\frac{10!}{(2!)(2!)}$.

Because this is not a free tutorial site, that is all you get.

5. ## Re: Hi! I am a Computer Science Engineer studying math for an exam

Originally Posted by Plato
Because this is not a free tutorial site, that is all you get.
That helps a lot!! Thank you! I think I understand the second part also now
Answer for the second part: since they do not change their relative positions we only change the vowels and consonants with each other (6 each)
$\frac{6!}{(2!)(3!)}*\frac{6!}{(2!)}$